# CBSE Class 12th Math 6 – Application of Derivatives MCQs

#### The radius of a sphere is found to be 7 cm. If an error of 0.01 cm is made in measuring the radius, then the error made in measuring the surface area of the sphere is _____________.

Correct! Wrong!

S = 4Π r2 Then, dS = 8Πr dr dr = 0.01 cm Approximate error in S = 8(22/7) (7)0 .01 = 1.76 square cm Hence, if the radius is 7 am and an error of 0.01 cm is made in measuring the radius, the error made in measuring the surface area of the sphere is 1.76 sq. cm.

#### The radius of a sphere is decreased from 10 cm to 9.8 cm. The approximate change in its volume is _________ (Use Π = 3.14)

Correct! Wrong!

We know V = Π r3 Therefore, dV/dr = 3 πr2 Or, dV = 3πr2dr Here, r = 10 cm and dr = -0.2 cm. Substituting in the equation, we get dV = 3 × π × 102 × (-0.2) = -188.4 cm3 The approximate change in the volume of the sphere is -188.4 cm2.

#### Let A ⊆ ℝ be a set, and let f: A → ℝ be a function. Then f(x) is said to be the function f is increasing if x < y implies f(x) ≤ f(y) for all x, y ∈ A.

Correct! Wrong!

Let A ⊆ ℝ be a set, and let f: A → ℝ be a function. Then f(x) is said to be the function f is increasing if x < y implies f(x) ≤ f(y) for all x, y ∈ A.

#### The nature of the function f(x) = -x/2 + sinx in the interval –π/3 < x < π/3 is _____________________.

Correct! Wrong!

The nature of the function f(x) = -x/2 + sinx in the interval –π/3 < x < π/3 is increasing. F(x) = -x/2 + sinx f'x = -1/2 + cosx Since –π/3 < x < π/3, 1/2 0. Therefore, f(x) is increasing in the interval.

#### The function f(x) = x3 - 3x + 8 has local maximum at _____________ .

Correct! Wrong!

Given f(x) = x3 - 3x + 8, f'(x) = 3 x3 – 3 The critical points are obtained by equating f'(x) = 0 3(x3 - 1) = 0: x = ±1 Applying the first derivative test, take two points on either side of -1, say -0.5 and -1.5. f'(x) at -1.5 = 3 × (2.25 - 1) is positive f'(x) at -0.5 is 3 × (0.25 - 1) is negative Sign changes from positive to negative means -1 is a point of maximum. Now, check for 1. Take 2 points, 0.5 and 1.5, on either side of 1. Sign of f'(x) at x = 0.5 = 3 × (0.25 - 1) is negative Sign of f'(x) at x = 1.5 = 3 × (2.25 - 1) is positive Sign changes from negative to positive means 1 is a point of minimum. Hence, the function f(x) = x3 - 3x + 8 has local maximum at x = -1.

#### The rate of a chemical reaction y, is given by the formula y = kx (a - x), where x is the amount of product, a is the amount of material at the beginning of the reaction and k > 0. The value of x for which the rate of chemical reaction is maximum is ____.

Correct! Wrong!

The value of x for which the rate of chemical reaction is maximum is a/2. f (x) = kx (a - x) Differentiating w.r.t x, f' (x) = k (a - 2x) For max or min, put f' (x) = 0 ⇒ x = a/2 f'' (a/2) = -2k < 0 ⇒ Maximum rate

#### The equation of the tangent to the curve y = 2x3 - x2 + 3 at (1, 4) is _______________.

Correct! Wrong!

The equation of the tangent to the curve y = 2x3 - x2 + 3 at (1, 4) is y - 4x = 0. Dy/dx = 6 x2 - 2x dy/dx(1, 4) = 6 - 2 = 4 The equation of the tangent is y - 4 = 4(x - 1) y - 4x = 0

#### The point on the curve y = x3 - 11x + 8, at which the tangent has the equation y = x - 8, is ___________.

Correct! Wrong!

y = x3 - 11x + 8 dy/dx = 3 x2 – 11 dy/dx = 3 x2- 11 Then, its slope is 1 dy/dx= 3 x2 - 11 = 1 3 x2= 12 x2 = 4 x = ± 2 When x = 2, y = 23 - 11(2) + 8 = -6 when x = -2, y = -23 - 11(-2) + 8 = 22 (2, -6) satisfies the tangent equation.

#### The interval in the function f(x) = sinx is strictly decreasing in ____.

Correct! Wrong!

Given, f(x) = sinx f'(x) = cosx Since for each, x ∈ (π/2, π), cosx < 0 We have, f'(x) < 0 Hence, f is strictly decreasting in (π/2, π).

#### If y = x2 + x - 2 and if x changes from 1 to 0.99, then the approximate change in y is given by _______.

Correct! Wrong!

Given x = 1 and δx = -0.01. From y = x2 + x - 2, we get, dy/dx = 2x + 1 dy = (2x + 1)dx = (2 + 1) × (-0.01) = 3 × (-0.01) Therefore, approximate change in y = -0.03. If y = x2 + x - 2 and if x changes from 1 to 0.99, then the approximate change in y is given by -0.03.

#### The function f(x) = x2 + 2x + 5 is a strictly decreasing for x < −1.

Correct! Wrong!

We have, f(x) = x2 + 2x + 5 f'(x) = 2x + 2 Now, f'(x) = 0 ⇒2x + 2 = 0 ⇒ x = −1 Point x = −1 divides the real line into two disjoint intervals. i.e. (−∞, −1) and (−1, ∞) In interval (−∞, −1), f'(x) < 0 ⇒ 2x + 2 < 0 Therefore, f is strictly decreasing in interval (−∞, −1) Hence, f is strictly decreasing for x < −1.

#### Let A ⊆ ℝ be a set, and let f : A → ℝ be a function. Then f(x) is said to be the function f is strictly increasing if x f(y) for all x, y ∈ A.

Correct! Wrong!

Let A ⊆ ℝ be a set, and let f: A → ℝ be a function. Then f(x) is said to be the function f is strictly increasing if x < y implies f(x) < f(y) for all x, y ∈ A.

#### Let A ⊆ ℝ be a set, and let f : A → ℝ be a function. Then f(x) is said to be the function f is decreasing if x < y implies f(x) ≥ f(y) for all x, y ∈ A.

Correct! Wrong!

Let A ⊆ ℝ be a set, and let f: A → ℝ be a function. Then f(x) is said to be the function f is decreasing if x < y implies f(x) ≥ f(y) for all x, y ∈ A.

#### One root of the equation x4 - 12x2 - 12x - 3 = 0 is approximately 4. A better approximation (upto 2 places of decimal) of the root is _______.

Correct! Wrong!

If x = a is an approximate solution of f(x) = 0 and a + δx is the correct solution, then δx = -f(a)f'(a). Better approximation is given by a –f(a)f'(a). Here, a = 4, then f(a) = f(4) = 44 - 12 × 42- 12 × 4 – 3 = 256 - 192 - 48 – 3 = 13 f'(x) = 4x3 - 24x – 12 f'(4) = 4 × 43 - 24 × 4 – 12 = 256 - 96 – 12 = 148 Better approximation is 4- (13/148) = 3 . 91 (upto 2 places of decimal) Hence, if one root of the equation x4 - 12x2 - 12x - 3 = 0 is approximately 4, a better approximation (upto 2 places of decimal) of the root is 3.91.

#### The function f(x) = cos 2x in (0, π/2) is ____________________.

Correct! Wrong!

The function f(x) = cos 2x in (0, π/2) is decreasing. f(x) = cos 2x f'(x) = -2 sin 2x 0 < x < π/2 0 < 2x 0, therefore -2 sin 2x < 0. Hence, it is decreasing.

#### Let A ⊆ ℝ be a set, and let f : A → ℝ be a function. Then f(x) is said to be the function f is strictly decreasing if x < y implies f(x) ≥ f(y) for all x, y ∈ A.

Correct! Wrong!

Let A ⊆ ℝ be a set, and let f: A → ℝ be a function. Then f(x) is said to be the function f is strictly decreasing if x < y implies f(x) ≥ f(y) for all x, y ∈ A.

#### The value of a for which the function f(x) = 2 x2 - 4ax + 9 is strictly increasing when x > 0 is ______.

Correct! Wrong!

f(x) = 2 x2 - 4ax + 9 f'(x) = 4x - 4a = 4(x - a) > 0 implies x > a a ≤ 0 The value of a for which the function f(x) = 2 x2 - 4ax + 9 is strictly increasing when x > 0 is ≤ 0.

#### The nature of the function f(x) = (1/2)x on R is ___________.

Correct! Wrong!

f(x) = (1/2)x f(x)=(1/2)xlog1/2 =f'x (1/2)x log1/2 = f'(x) = (1/2)x (-log2) < 0 is always decreasing. The nature of the function f(x) =(1/2)x is decreasing.

#### The least value of ex + e(-x) is ____.

Correct! Wrong!

The least value of ex + e(-x) is 2. f' (x) = ex - e(-x) f'' (x) = ex + e(-x) Put f' (x) = 0 ⇒ x = 0 f'' (0) = 2 > 0 Minimum value of f (x) = 2 at x = 0.

#### A ladder 13 m long is leaning against a wall. The foot of the ladder is pulled along the ground towards the wall at the rate of 3 cm/s. Then, the rate at which its height is increasing when the foot of the ladder is at a distance of 5 m from the wall is _______.

Correct! Wrong!

From the Pythagoras equation, we know that x2 + y2 = 132. Therefore, 2x (dx/dt) + 2y(dy/dt) = 0 From the equation x2 + y2 = 132, when x = 5, y = 12. dx/dt = -3, since it is pulled towards the wall. Then dy/dt is positive since the height increases. 5 (-3) + 12 dy/dt = 0 dy/dt = 15/12 = 5/4 cm/s

#### Let x and y be positive variables related by the equation x y2 = 16. If x is measured as 1 ± 0.001, then the value of y is _________.

Correct! Wrong!

x y2 = 16 x = 1 and δx = ± 0.001 Substituting x = 1 in x y2 = 16, we get y2 = 16 ⇒ y = 4 Differentiating x y2 = 16 we get, 2xy dy/dx + y2 = 0 Therefore, dy/dx = - y/2x Approximate error in y is given by dy = (dy/dx) δx = -42 (±0 .001) = ∓ 0 .002 y = 4 ∓ 0 .002 By the equation x y2 = 16, if x is measured as 1 ± 0.001, then y = 4 ∓ 0 .002.

#### For all values of a and b, f(x) = x3 + 3ax2 + 3a2x + 3a3 + b is increasing.

Correct! Wrong!

Given, f(x) = x3 + 3ax2 + 3a2x + 3a3 + b f'(x) = 3x2 + 6ax + 3a2 = 3(x2 + 2ax + a2) = 3(x + a)2≥ 0 Hence, for all values of a, b the function f is increasing.

#### The length l of a rectangle is decreasing at the rate of 3 cm/s and its breadth b is increasing at the rate of 2 cm/s. Then, the rate of change of area when l = 20 m and b = 10 m is ________.

Correct! Wrong!

Area of a rectangle is = length × breadth A = l × b dA/dt = l (db/dt) + b (dl/dt) = 2000 × (-3) + 1000 × (2 = -4000 cm2/s

#### The function f(x) = 3x + 13 is strictly increasing on ℝ.

Correct! Wrong!

Let x1 and x2 be any two numbers on ℝ. Then, x1 < x2 ⇒ 3x1 < 3x2 ⇒ 3x1 + 13 < 3x2+ 13 ⇒ f(x1) < f(x2) Hence, f is strictly increasing on ℝ.

#### f (x) = |x| has ____.

Correct! Wrong!

f (x) = |x| has minimum at x = 0. f (x) = |x| > = 0 Minimum value of f (x) = 0 at x = 0

CBSE Class 12th Math 6 - Application of Derivatives MCQs
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